Question

1．巧解下列關(guān)于x的方程（組）：$\left\{\begin{array}{l}{\frac{3}{x+y}+\frac{10}{x-y}=-5}\\{\frac{2}{x+y}-\frac{15}{x-y}=1}\end{array}\right.$．

Answer 1

分析 根據(jù)．換元法，可得方程的解

解答 解：設(shè)$\frac{1}{x+y}$=u，$\frac{1}{x-y}$=v，
原方程等價于$\left\{\begin{array}{l}{3u+10v=-5①}\\{2u-15v=1②}\end{array}\right.$，
解得$\left\{\begin{array}{l}{u=-1}\\{v=-\frac{1}{5}}\end{array}\right.$，
$\left\{\begin{array}{l}{x+y=-1}\\{x-y=-5}\end{array}\right.$，
解得$\left\{\begin{array}{l}{x=-3}\\{y=2}\end{array}\right.$．

點評 本題考查了解二元一次方程組，利用換元法是解題關(guān)鍵．